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Suppose $\sum_{n = 0}^\infty a_n$ converges. Can we conclude (from the definition of convergence), that for every $\epsilon > 0$, there exists a positive integer $N$ such that $k > N$ implies $|\sum_{n = k+1}^\infty a_n| < \epsilon$?

Old Question

We say a series $\sum_{n = 0}^\infty a_n$ converges (assume $a_n$ are real), if the sequence of partial sums converges. That is, there exists a real number $l$ such that for every $\epsilon > 0$, there exists a positive integer $N$ such that $k >N$ implies $| \sum_{n = 1}^k a_n - l | < \epsilon$. My question is this definition equivalent to saying for every $\epsilon > 0$, there exists $N$ such that $k > N$ implies

$$\bigg| \sum_{n = k + 1}^\infty a_n\bigg| < \epsilon$$

This should be obvious if we take $|\sum_{n = 0}^\infty a_n - \sum_{n = 0}^k a_n|$ to be the above value, but I'm uncomfortable substracting term by term, since technically we're not really "adding up all terms".

Can someone clear this up?

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The problem is that you are trying to define what $\sum$ means for an infinite number of terms. Until you define the notion of convergence, you must stick with finite sums. –  copper.hat Apr 1 '13 at 19:41
@copper.hat I rewrote the question. My original question was not the one I had in mind. –  user70483 Apr 1 '13 at 19:47
Yes, if the series converges, all of its tail sums converge, and the sequence of tail sums converges to $0$. –  Brian M. Scott Apr 1 '13 at 19:48
@BrianM.Scott My whole question rests on this point. If $\sum_{n=0}^\infty a_n$ converges is the operation $\sum_{n=0}^\infty a_n - \sum_{n=0}^k a_n = \sum_{n=k+1}^\infty a_n$ valid? –  user70483 Apr 1 '13 at 19:49
@user70483 Yes, it is. –  Pedro Tamaroff Apr 1 '13 at 21:08

2 Answers 2

Answer to new question:

It seems a matter of notation only. By using $$\sum_{n={k+1}}^\infty a_n$$ you're just obscuring the fact this is $$\sum_{n=0}^\infty a_n-\sum_{n=0}^k a_n$$

We write $\displaystyle \sum_{n=0}^\infty a_n$ to denote the limit of the sequence of partials sums. Thus, if we say that $\langle a_n\rangle $ is summable, we say that for each $\epsilon >0$ there exists an $N$ such that $$\tag 1\left| \sum_{n=0}^\infty a_n-\sum_{n=0}^k a_n\right|<\epsilon$$

whenever $k>N$. On the other hand, we could as well say, for fixed $k$, that $$\sum_{n={k+1}}^\infty a_n$$ is the limit of the partial sums of $a_k,a_{k+1},\dots$. In any case, it is almost tautological. We're just repeating yet obscuring the definition in $(1)$.

We cannot start talking about $$\sum_{k=m}^\infty$$ unles we know something about the convergence a priori. What we can do is use Abel's criterion

PROP A sequence $\langle a_n:n\in\Bbb N\rangle$ is summable if given any $p$ and $\epsilon >0$, there exists $N\in\Bbb N$ such that for each $n>N$, we have that $$\sum_{k=n}^{n+p}a_k<\epsilon$$

Alternatively, we have the Cauchy criterion

PROP A sequence $\langle a_n:n\in\Bbb N\rangle$ is summable if for every $\epsilon >0$ there exists $N\in\Bbb N$ such that $$\sum_{k=m+1}^na_k<\epsilon$$ whenever $m,n>N$:

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I apologize, but my original was completely wrong. I rewrote my question. –  user70483 Apr 1 '13 at 19:48
@user70483 It is very bad practice to completely rewrite a question after it has been answered. I will leave both of your questions displayed, and answer the new one in a while. –  Pedro Tamaroff Apr 1 '13 at 19:49

The answer is yes, to see this consider the sequence of partial sums $$s_n=\sum_{k=1}^{n}a_k$$ and assume that $s_n$ converges to $s=\sum_{k=1}^{\infty}a_k$ as $n\to\infty.$ That means for any $\epsilon>0$ there exists $N\in \Bbb N$ such that $$|s_n-s|<\epsilon$$ for $n\ge N $, but we have for $n\ge N$$$\begin{align} |s_n-s|&=\bigg|\sum_{k=1}^{n}a_k-\sum_{k=1}^{\infty}a_k\bigg|\\&=\bigg|\sum_{k=n+1}^{\infty}a_k\bigg|<\epsilon \end{align}$$Thus, the result.

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